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TEA [102]
2 years ago
6

These box plots show daily low temeratures for a samle of days in two different towns

Mathematics
1 answer:
Zarrin [17]2 years ago
4 0

The correct statement about the center of both box plots is the median for Town A, 30° is greater median for Town B, 25°

<h3>What is the correct statement?</h3>

A box plot is used to study the distribution of a dataset. The box plot consists of two lines and a box. The two lines are known as whiskers. The whiskers represent the minimum and maximum numbers.

On the box, the first line to the left represents the lower (first) quartile. The next line on the box represents the median. The third line on the box represents the upper (third) quartile.

Please find attached the complete question. To learn more about box plots, please check: brainly.com/question/27215146

#SPJ1

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In the figure given, what is the value of x?
torisob [31]

Answer:

x = 30

Step-by-step explanation:

In the figure attached,

Two lines are parallel and one line having two adjacent angles (4x)° and (x + 30)° are the transverse.

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4 0
3 years ago
Many smartphones, especially those of the LTE-enabled persuasion, have earned a bad rap for exceptionally bad battery life. Batt
BartSMP [9]

Answer:

Null hypothesis: the variance in hours of usage for talking is not greater than the the variance in hours of usage for internet.

H_o: \sigma _1 ^2 \leq \sigma _2^2

Alternative hypothesis: the variance  in hours of usage for talking is  greater than the the variance in hours of usage for internet.

H_a : \sigma_1^2 > \sigma_2^2

\mathbf{s_ 1 =16.11}

\mathbf{s_2 = 7.98}

Step-by-step explanation:

Let x_1 and x_2 be the two variables that represents the battery life in hours for talking usage and battery life in hours for internet usage respectively.

The hypothesis can be formulated as:

Null hypothesis: the variance in hours of usage for talking is not greater than the the variance in hours of usage for internet.

H_o: \sigma _1 ^2 \leq \sigma _2^2

Alternative hypothesis: the variance  in hours of usage for talking is  greater than the the variance in hours of usage for internet.

H_a : \sigma_1^2 > \sigma_2^2

The standard deviation for the battery usage for talking is :

\bar x_1 = \dfrac{1}{n_1} \sum x_i  \\ \\ \bar x_1  = \dfrac{1}{12}(35.8 +22.4+...+35.5) \\ \\ \bar x_1 = \dfrac{241.2}{12}  \\ \\ \bar x_1 =20.1

The standard deviation Is:

s_ 1 = \sqrt{\dfrac{1}{n_1-1}\sum (x{_1i}-\bar x_i)^2}

s_ 1 = \sqrt{\dfrac{1}{12-1}\sum (35.8- 20.1)^2+ (35.5-20.1)^2}

s_ 1 = \sqrt{259.568}

\mathbf{s_ 1 =16.11}

The standard deviation for the battery life usage for the internet is :

\bar x_2 = \dfrac{1}{n_2} \sum x_{2i}

\bar x_2 = \dfrac{1}{10} (24.0+12.5+36.4+...+4.7})

\bar x_2 = \dfrac{115}{10}

\bar x_2 = 11.5

Thus; the standard deviation is:

s_2 = \sqrt{\dfrac{1}{n_2-1}(x_{2i}- \bar x_2)^2}

s_2 = \sqrt{\dfrac{1}{10-1}(24-11.5)^2+(4.7-11.5)^2}

s_2 = \sqrt{63.60}

\mathbf{s_2 = 7.98}

4 0
3 years ago
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